Equal determinants

Let  A, B \in \mathbb{R}^{n \times n} that are diagonizable in  \mathbb{R} . If  \det (A^2+B^2)=0 and  AB=BA   then prove that

 \det A = \det B =0

Solution

The key point here is that the matrices are simultaneously diagonisable. Thus , there exists an invertible matrix C\in\mathbb{M}_{n}(\mathbb{R}) and diagonisable matrices P\,,Q\in\mathbb{M}_{n}(\mathbb{R}) such that

A=C\,P\,C^{-1}\,\,,B=C\,Q\,C^{-1}

Hence

    \begin{align*} 0&=\det(A^2+B^2)\\&=\det(C\,P^2\,C^{-1}+C\,Q^2\,C^{-1})\\ &=\det(C\,(P^2+Q^2)\,C^{-1})\\ &=\det(P^2+Q^2)\\ &=\prod_{i=1}^{n}\left(p_{i}^2+q_{i}^2\right) \quad (1) \end{align*}

where p_{i}\,,1\leq i\leq n are the diagonial elements of P and q_{i}\,,1\leq i\leq n of Q. According to (1) we have that

p_{i}^2+q_{i}^2=0 \quad \text{for some} \; i\in\left\{1,...,n\right\}

But p_{i}\,,q_{i}\in\mathbb{R}, so p_{i}=q_{i}=0 and finally we conclude that:

\displaystyle \det(A)=\det (P)=\prod_{i=1}^{n}p_{i}=0=\prod_{i=1}^{n}q_{i}=\det(Q)=\det(B)

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Hadamard Inequality

Let \mathbf{a_1, a_2, \dots, a_N} be column vectors in \mathbb{R}^N and let A=(a_1, a_2, \dots, a_N) be the corresponding N \times N real matrix. Then the following inequality holds:

(1)   \begin{equation*} \left | \det A \right |\leq \prod_{n=1}^{N}\left \| a_n \right \| \end{equation*}

where \left \| \cdot \right \| is the Euclidean norm on vectors in \mathbb{R}^N. In continuity , give the geometrical interpretation of the inequality above.

Solution

By the Gramm – Schmidt process we can establish the existence of an orthonormal basis \mathbf{b_1, b_2, \dots, b_N}$ for $\mathbb{R}^N such that

(2)   \begin{equation*} {\rm span}_{\mathbb{R}} \left \{ \mathbf{a_1, a_2, \dots, a_N} \right \}={\rm span}_{\mathbb{R}}\left \{ \mathbf{a_1, a_2, \dots, b_N} \right \} \end{equation*}

for each n=1, 2, \dots, N. Now, we may write B=(\mathbf{b_1, b_2, \dots, b_N}) for the corresponding N \times N real and orthogonal matrix. By orthogonality each vector \xi in \mathbb{R}^N has an expansion as:

\displaystyle \xi = \sum_{n=1}^{N}\langle \xi, \mathbf{b}_n\rangle \mathbf{b}_n \Rightarrow \left \| \xi \right \|^2 = \sum_{n=1}^{N}\left | \langle \xi, \mathbf{b}_n \rangle \right |^2

On the other hand (2) implies that each vector \mathbf{a}_m has a shorter expansion of the form:

(3)   \begin{equation*} \mathbf{a}_m = \sum_{n=1}^{m}\langle \mathbf{a}_m, \mathbf{b}_n \rangle \mathbf{b}_n \end{equation*}

Alternatively let C=(c_{k\ell}) be the N \times N upper triangular matrix defined as:

\displaystyle c_{k\ell}= \langle \mathbf{a}_\ell, \mathbf{b}_k \rangle \;\; \text{if} \; 1 \leq k \leq \ell \;\; \textbf{and} \;\; c_{k \ell}=0 \; \; \text{if} \; \ell < k \leq N

Then (3) is restated as A=BC and using again the fact that B has orthonormal columns and the fact that C is upper triangular we get:

    \begin{align*} \left ( \det A \right )^2 &=\det \left ( A^T A \right ) \\ &= \det \left ( C^T B^T BC \right )\\ &= \det \left ( C^T C \right )\\ &= \left ( \det C \right )^2 \\ &= \prod_{n=1}^{N}\left | \langle \mathbf{a}_n , \mathbf{b}_n \rangle \right |^2 \\ &\leq \prod_{n=1}^{N} \left ( \sum_{m=1}^{n}\left | \langle \mathbf{a}_n, \mathbf{b}_n \rangle \right |^2 \right ) \\ &=\prod_{n=1}^{N} \left \| \mathbf{a}_n \right \|^2 \end{align*}

Notes:

  • The above argument also shows that there exists equality if and only if

    \displaystyle \left | \langle \mathbf{a}_n, \mathbf{b}_n \rangle \right |^2 = \sum_{m=1}^{n}\left | \langle \mathbf{a}_m , \mathbf{b}_n \rangle \right |^2

    for each n. That is , if and only if, \mathbf{a}_n= \langle \mathbf{a}_n, \mathbf{b}_n \rangle \mathbf{b}_n. This can only be achieved if the vectors \mathbf{a}_n are pairwise orthogonal.

  • The geometrical interpretation of this inequality is the following: The volume of an n dimensional parallelepiped produced by n vectors can not exceed the product of their measures.

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No rational function

Prove that there exists no rational function such that

\displaystyle f(n)=1+ \frac{1}{2} + \cdots + \frac{1}{n} \quad \text{forall} \; n \in \mathbb{N}

Solution

Suppose , on the contrary , that such function exists. Since the harmonic series diverges we conclude that the limit of our function in infinity is infinity. This, in return means that the degree of the nominator , call that m is greater that the one of the denominator , call that n. Extracting x^{n-m} in the nominator we get that

\displaystyle R(x)=\frac{P(x)}{Q(x)} = x^{m-n} s(x)

The limit of s at infinity is finite and call that \ell. Hence:

\begin{aligned} \lim_{x \rightarrow +\infty} \left ( \frac{f(x)}{g(x)} - \ln x \right ) &= \lim_{x \rightarrow +\infty} \left [ x^{m-n} s(x) - \ln x \right ] \\ &= \lim_{n \rightarrow +\infty} x^{m-n} \left [ s(x) - \frac{\ln x}{x^{m-n}} \right ]\\ &= +\infty \end{aligned}

and of course this contradicts the fact that

\displaystyle \lim \left ( \mathcal{H}_n - \ln n \right ) = \gamma

where \gamma is the Euler  – Mascheroni constant .

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Non existence of sequence of continuous functions

Prove that there does not exist a sequence of continuous functions f_n:[0, 1] \rightarrow \mathbb{R} such that converges pointwise, to the function \chi_{\mathbb{Q}} , where \chi_{\mathbb{Q}} is the characteristic polynomial of the rationals in [0, 1].

Solution

The indicator of the rationals is no other function than

\chi_{\mathbb{Q}}= \left\{\begin{matrix} 1& ,& x \in \mathbb{Q}\\ 0& , & \text{elsewhere} \end{matrix}\right.

It is known that pointwise limits of continuous functions have a meagre set of points of continuity. However, this function is discontinuous everywhere and thus we cannot expect a sequence of continuous functions to converge pointwise to it.

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On a determinant

Let p be a prime number and let \omega be  a primitive p-th root of unity. Define:

\mathcal{V} = \det \begin{pmatrix} 1 &1  &1  &\cdots  &1 \\  1& \omega &\omega^2  &\cdots  &\omega^{p-1} \\ 1 &\omega^2  &\left ( \omega^2 \right )^2  &\cdots  &\left ( \omega^2 \right )^{p-1} \\  \vdots&\vdots  &\vdots  &\ddots  & \vdots\\  1& \omega^{p-1} &\left ( \omega^{p-1} \right )^2  &\cdots  & \left ( \omega^{p-1} \right )^{p-1} \end{pmatrix}

Evaluate the rational number \mathcal{V}^2.

Solution

The ij -th entry of \mathcal{V} is  \omega^{(i-1)(j-1)}. Thus the ij-th entry of \mathcal{V}^2 is equal to:

\begin{aligned} \sum_{\ell} \omega^{(i-1)(\ell-1)} \omega^{(\ell-1) (j-1)} &= \sum_{\ell} \omega^{(i-1+j-1)(\ell-1)} \\ &= \left\{\begin{matrix} 0 &\text{if} & (i-1) + (j-1) \neq 0 \mod p\\ p& \text{if} & (i-1) + (j-1) = 0 \mod p \end{matrix}\right. \\ \end{aligned}

since it is known that \sum \limits_{0 \leq \ell <p} \omega^{\ell}=0 for any p-th root of unity \omega rather than 1. Thus:

\mathcal{V}^2 = \begin{pmatrix} p &0 & 0 &\cdots &0 &0 \\ 0 & 0 & 0 & \cdots & 0 & p\\ 0& 0 & 0 &\cdots & p &0 \\ \vdots& \vdots &\vdots &\ddots &\vdots &\vdots \\ 0& 0 & p& \cdots&0 & 0\\ 0& p & 0 &\cdots &0 &0 \end{pmatrix}

Thus is there is a p at the upper left corner and p ‘s along the anti diagonal in the lower right (n-1) \times (n-1) block. Thus:

\left ( \det \mathcal{V} \right )^2 = \det \left ( \mathcal{V}^2 \right )= (-1)^{(p-1)(p-2)/2} \; p^p

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